A horizontal, uniform bar of mass m = 7.97 kg and length L = 3.5 m is suspended
ID: 1463507 • Letter: A
Question
A horizontal, uniform bar of mass m = 7.97 kg and length L = 3.5 m is suspended by two vertical strings. String 1 is attached to one end of the bar, string 2 is attached a distance L/3 in from the other end. A monkey of mass m/2 moves along the bar, starting from the end attached to string 1. Eventually you will be asked to find the tension in the strings when the monkey is in the middle of the bar.
A) Which of the free body diagrams correctly represents the forces applied to the bar when the monkey is in the centre? E.g., enter A.
B)
Which of the following statements are true? E.g., if A and C are true, enter AC; if none are true, enter N. You only have 5 tries!
The sum of the tensions T1 and T2 equals the weight of the bar and the monkey.
The bar should be modelled as a rigid rod in static equilibrium.
Choosing the pivot point to be where string 1 attaches to the bar would lead to a torque equation where the tension T2 would not appear.
When the monkey reaches the middle of the bar, it is a distance L/3 from string 2.
The sum of the forces on the bar must be zero.
C) What is the tension, T1, in string 1?
D )What is the tension, T2, in string 2?
E)
The bar remains horizontal when the monkey moves past string 2 to the end of the bar.
Enter T for true, F for false, or C for cannot tell. You have only one try!
Explanation / Answer
(A) - Correct Option - B
(B)
The sum of the tensions T1 and T2 equals the weight of the bar and the monkey - True
The bar should be modelled as a rigid rod in static equilibrium - True.
The sum of the forces on the bar must be zero - True.
(C)
The sum of the tensions T1 and T2 equals the weight of the bar and the monkey =
T1 + T2 = mg + mg/2
T1 + T2 = 7.97 * 9.8 + 7.97/2 * 9.8 N
T1 + T2 = 117.15 N
Let us choose the pivot point , Where string 1 is attached -
The sum of the forces on the bar must be zero =
Therefore,
(mg + mg/2) * L/2 = T2 * (L - L/3)
(mg + mg/2) * L/2 = T2 * 2*L/3
117.15/2 = T2 * 2/3
T2 = 87.86 N
T1 = 117.15 - 87.86 N
T1 = 29.3 N
(C) Tension T1 in string, T1 = 29.3 N
(D) Tension T2 in string, T2 = 87.86 N
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